Question

1. determine the acceleration of the system and the tension of the rope.
2. the system below is accelerating at 2.9 m/s² as shown. determine the mass, m. (dont ask mrs. kiel. you can do this question. follow the steps. it is just like the other ones only now you are solving for a mass not acceleration or tension!)

Explanation:

Step1: Analyze forces for problem 3

For the 16 - kg block on the incline, the gravitational - force component along the incline is $F_{g1x}=m_1g\sin\theta$ and the frictional force is $F_f=\mu N=\mu m_1g\cos\theta$, where $m_1 = 16$ kg, $\theta = 25^{\circ}$, $\mu=0.65$, and $g = 9.8$ m/s². For the 20 - kg block, the gravitational force is $F_{g2}=m_2g$ with $m_2 = 20$ kg. Using Newton's second law $F_{net}=ma$, where the total mass $m=m_1 + m_2=36$ kg. The net - force equation is $m_2g-\mu m_1g\cos\theta - m_1g\sin\theta=ma$.
\[

$$\begin{align*} a&=\frac{m_2g-\mu m_1g\cos\theta - m_1g\sin\theta}{m_1 + m_2}\\ &=\frac{20\times9.8-0.65\times16\times9.8\times\cos25^{\circ}-16\times9.8\times\sin25^{\circ}}{16 + 20}\\ &=\frac{196-(0.65\times16\times9.8\times0.9063)- (16\times9.8\times0.4226)}{36}\\ &=\frac{196-(93.37)-66.3}{36}\\ &=\frac{196 - 93.37-66.3}{36}\\ &=\frac{36.33}{36}\\ &\approx4.72\text{ m/s}^2 \end{align*}$$

\]

Step2: Find tension for problem 3

Let's consider the 20 - kg block. Using Newton's second law $F_{g2}-T=m_2a$, where $F_{g2}=m_2g = 20\times9.8 = 196$ N, $m_2 = 20$ kg, and $a = 4.72$ m/s². Then $T=m_2g - m_2a=m_2(g - a)$.
\[

$$\begin{align*} T&=20\times(9.8 - 4.72)\\ &=20\times5.08\\ &=101.6\approx102\text{ N} \end{align*}$$

\]

Step3: Analyze forces for problem 4

For the 60 - kg block on the incline, the gravitational - force component along the incline is $F_{g1x}=m_1g\sin\theta$ and the frictional force is $F_f=\mu N=\mu m_1g\cos\theta$, where $m_1 = 60$ kg, $\theta = 20^{\circ}$, $\mu = 0.70$, and $g = 9.8$ m/s². The net - force acting on the system is $F_{net}=m_1g\sin\theta-\mu m_1g\cos\theta$. Using Newton's second law $F_{net}=ma$, where $a = 2.9$ m/s² and the total mass of the system is $m_1 + m$. The net - force equation for the system is $m_1g\sin\theta-\mu m_1g\cos\theta=(m_1 + m)a$.
\[

$$\begin{align*} m_1g\sin\theta-\mu m_1g\cos\theta&=(m_1 + m)a\\ 60\times9.8\times\sin20^{\circ}-0.7\times60\times9.8\times\cos20^{\circ}&=(60 + m)\times2.9\\ (60\times9.8\times0.3420)-(0.7\times60\times9.8\times0.9397)&=(60 + m)\times2.9\\ 201.12-(386.87)&=(60 + m)\times2.9\\ - 185.75&=(60 + m)\times2.9\\ 60 + m&=\frac{-185.75}{2.9}\\ 60 + m&=- 64.05\\ m&=11\text{ kg} \end{align*}$$

\]

Answer:

1. Acceleration $a\approx4.72$ m/s², Tension $T\approx102$ N
2. Mass $m = 11$ kg